Solving Systems of Linear Equations by Addition and in Three Variables
Quiz yourself by thinking what should be in
each of the black spaces below before clicking
on it to display the answer.
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Solve the System: {x+y=1 (1) {x-y=5 (2) | Add equations (1)+(2)to eliminate y. x=3
Solve for y. By plugging in x to either (1) or (2) y=-2
Solution is (3,-2)
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Solve the System: {-2x+y=4 (1) {-x+3y=-8 (2) | Before adding equations must distribute -3 to (1) to eliminate y when adding (1)+(2), x=2.857 Solve for y. by plugging in x in either (1) or (2) y= 1.714
Solution is (2.857,-1.714)
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Solve the System: {4x-2y=4 (1) {-8x+4y=-8 (2) | Before adding both equations must distribute 2 to (1) to eliminate y when adding (1)+(2), Solve for x. When solving for x everything ends up canceling out so the answer is 0=0 meaning infinite number of solutions.
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Solve the System: {-x-(y/2)=7/2 (1) {(-x/2)+(y/4)=0 (2) | First must eliminate fractions, multiply (1) by 2, mulftiply (2) by 4. Add equations
x=-4/7 Plug in x in either (1) or (2) to solve for y. y=-5.857
Solution is (-4/7,-5.857)
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Solve the System: {2x-y+z=-8 (1) {x+2y-z=2 (2){2x+4y-2z=0 (3) | Add (1)+(2) eliminate z becomes (4)
Then add (1)+(3) to eliminate z again becomes (5) Now (4)+(5) to eliminate y,solve for x x=-7/3 Replace x in(4)or(5) solve for, y=1 Finally replace x and y (1)(2)or(3) solve for z.
Solution-(-7/3,1,7.3)
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Solve the System: {2x-4y+8z=2 (1) {-x-3y+z=11 (2) {x-2y+4z=0 | Add (2)+(3) to eliminate x, becomes (4) Eliminate x again by (1)+(2) becomes (5) Using (4) and (5)solve for y and z. when you do this you end up getting the answer 0=2 This means the system is false and has no solution.
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Solve the System: {4x+8y =1 (1) {-4x -4z=-1 (2) { y-4z=-3 (3) | Since there is no y in (2) add (1)+(3) to eliminate y. becomes equation 4. Next add (4)+(2) to solve for z. z=.893 plug that in to (3) and then solve for y. y=-.571 then use y to solve for x using equation (1). x=-.892
Solution is (.892,-.571,.893)
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Solve the System: {x-y+z=-4 (1) {3x+2y-z=5 (2) {-2x+3y-z=15 (3) | Add (1)+(2) to eliminate z. becomes (4) Then add (1)+(3) to eliminate z again. becomes (5). add (4)and(5) to eliminate y,solve for x, x= -1. then use (4)or(5) to solve for y y=5. Then replace x,y in (1)(2)or(3) to solve for z. z=2
Solution is (-1,5,2
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Solve the system: {2x-4y+6z+-8 (1) {x-5y+5z=3 (2) {3x+y-3z=5 (3) | Add (1)+(2) to eliminate z. becomes(4) then add (1)+(3) to eliminate z. becomes (5) add(4)+(5) to eliminate y,solve for x. x=3 then solve for y using (4)or(5) y=2, use (1)(2)or(3) to solve for z using x,y. z=2
soultion (3,2,2)
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Solve the system: {x+y+z=8 (1) {2x-y-z=10 (2) {x-2y-3z=22 (3) | Add (1)+(2) to eliminate z. becomes (4) then add (1)+(3) to eliminate z, becomes (5)Next add (4)+(5) to eliminate y & solve for x, x=6 replace x in (4) or (5) and solve for y. y=22. Then use (1)(2)(3) to solve for z using x,y z=-20
Solution: (6,22,-20)
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